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For a direction $e$, the Kakeya maximal function is the supremum over translations of normalized averages of $|f|$ over unit-length Kakeya tubes in direction $e$. It measures how large a function can be on at least one thin tube in each direction. Normalizing by tube volume is essential when comparing different thicknesses.

For a unit direction $e$, let $T_{\delta }(a,e)$ be a length-one Kakeya tube with transverse radius $\delta$, centered at $a$, and define the Kakeya maximal function byUsing normalized surface measure on the direction sphere, the Kakeya maximal conjecture is the following family of estimates, in the formulation relevant to this paper:The constant is independent of $\delta$ and $f$. Replacing round Kakeya tubes by comparable rectangular tubes changes only dimensional constants.

A bounded Kakeya set contains a unit line segment in every direction. The Kakeya Minkowski dimension conjecture says that every such set has full Minkowski dimension $n$. The maximal estimate in fact gives full lower as well as upper Minkowski dimension.

To prove that implication, let $E_{\delta }=\{x:\mathrm{dist}(x,E)<\delta \}$. Each unit segment in $E$ has a thinner tube contained in $E_{\delta }$, so ${\mathcal{K}}_{c\delta }{1}_{E_{\delta }}(e)\ge 1$ for every $e$, with a fixed dimensional $c>0$. Apply the maximal estimate to this indicator function:If $N_{\delta }(E)$ is the smallest number of radius-$\delta$ balls covering $E$, that cover, enlarged by a fixed factor, covers $E_{\delta }$. Thus $|E_{\delta }|{\lesssim }_n{\delta }^n{N}_{\delta }(E)$ andTaking the lower limit of $\log N_{\delta }(E)/\log (1/\delta )$ and then letting $\epsilon \downarrow 0$ gives lower Minkowski dimension at least $n$. Bounded subsets of ${\mathbb{R}}^n$ have upper Minkowski dimension at most $n$. Consequently both dimensions equal $n$, which proves the requested implication.

It is enough to prove the stronger planar Kakeya maximal function boundWe first work with a $\delta$-separated net of unoriented directions $e_1,\dots ,e_M$, with $M\asymp {\delta }^{-1}$. Write ${\alpha }_{ij}$ for angular distance modulo $\pi$. Choose arbitrary length-one, width-$\delta$ rectangles $T_j$ in these directions. The permitted rectangle intersection fact isThe first alternative includes parallel rectangles. Their locations are arbitrary; only separation of their directions matters.

Define $Af(j)={\delta }^{-1}{\int }_{T_j}f$, and put $\Vert a{\Vert }_{{\ell }_{\delta }^2}^2=\delta {\sum }_j|a_j{|}^2$. With these weights the adjoint operator is $A^{\ast }a={\sum }_j{a}_j{1}_{T_j}$. Since a separated angular net has only a bounded number of directions at each distance scale $k\delta$ from $e_i$, its overlap matrix satisfiesThe diagonal term is of size $\delta$. Using $|a_i{a}_j|\le (|a_i{|}^2+|a_j{|}^2)/2$ and symmetry gives the Schur test estimateBy duality of Lp spaces, $\Vert Af{\Vert }_{{\ell }_{\delta }^2}\lesssim \sqrt{\log (2/\delta )}\Vert f{\Vert }_2$. This is uniform over every choice of the translated rectangles. For each direction choose a rectangle approaching the supremum for $|f|$, then take the limit. That gives the same estimate for the discretized Kakeya maximal function.

To recover all directions, partition the direction circle into arcs of length comparable to $\delta$, each with a net direction. A tube in an arc is contained in a rectangle in its net direction with width $C\delta$ and length at most two. A bounded subdivision in the length direction reduces this to the same averaging operators; the wider tubes obey the same overlap estimate with fixed-factor changes. ThereforeFinally $\sqrt{\log (2/\delta )}\le C_{\epsilon }{\delta }^{-\epsilon }$ for every $\epsilon >0$. The planar Kakeya maximal estimate therefore has the required arbitrary small power loss.